[Toán] Chứng minh bất đẳng thức

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hunghinh2000

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chonhoi110

Đặt $\sqrt{3+\sqrt{3+\sqrt{3...+\sqrt{3}}}}=n$ (2015 dấu căn)

$\Longrightarrow n^2-3=\sqrt{3+\sqrt{3+\sqrt{3...+\sqrt{3}}}}$ (2014 dấu căn)

$\Longrightarrow VT=\dfrac{3-n}{6-(n^2-3)}=\dfrac{1}{n+3}$

Ta có: $1<n \Longrightarrow \dfrac{1}{n+3}<\dfrac{1}{1+3}=\dfrac{1}{4}$
 
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